PHYSICS COLLEGE

While a roofer is working on a roof that slants at 37.0 ∘∘ above the horizontal, he accidentally nudges his 92.0 NN toolbox, causing it to start sliding downward, starting from rest. If it starts 4.25 m from the lower edge of the roof, how fast will the toolbox be moving just as it reaches the edge of the roof if the kinetic friction force on it is 22.0 N?

Answers

Answer 1

Answer:

Answer:

The speed is $v =8.17 m/s$

Explanation:

From the question we are told that

The angle of slant is $\theta = 37.0^o$

The weight of the toolbox is $W_t = 92.0N$

The mass of the toolbox is $m = (92)/(9.8) = 9.286kg$

The start point is $d = 4.25m$ from lower edge of roof

The kinetic frictional force is $F_f = 22.0N$

Generally the net work done on this tool box can be mathematically represented as

$Net \ work done = Workdone \ due \ to \ Weight + Workdone \ due \ to \ Friction$

The workdone due to weigh is = $mgsin \theta * d$

The workdone due to friction is = $F_f \ cos\theta * d$

Substituting this into the equation for net workdone

$W_(net) = mgsin\theta * d + F_f \ cos \theta *d$

Substituting values

$W_(net) = 92 * sin (37) * 4.25 + 22 cos (37) * 4.25$

$= 309.98 J$

According to work energy theorem

$W_(net) = \Delta Kinetic \ Energy$

$W_(net) = (1)/(2) m (v - u)^2$

From the question we are told that it started from rest so u = 0 m/s

$W_(net) = (1)/(2) * m v^2$

Making v the subject

$v = \sqrt{(2 W_(net))/(m) }$

Substituting value

$v = \sqrt{(2 * 309.98)/(9.286) }$

$v =8.17 m/s$

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A 1000-kg car rolling on a smooth horizontal surface ( no friction) has speed of 20 m/s when it strikes a horizontal spring and is brought to rest in a distance of 2 m What is the spring’s stiffness constant?

Answers

Explanation:

kinetic energy was converted to potential energy in the spring.

the answer is in the above image

As the moon orbits the Earth which of the following changes (1) a. Speed b. Velocity c. Acceleration d. A, B, and C e. None

Answers

Answer:

B- Velocity

Explanation:

This means gravity makes the Moon accelerate all the time, even though its speed remains constant.

Determine whether the following statements are true and give an explanation or counterexample.(A) If the acceleration of an object remains constant, its velocity is constant.
(B) If the acceleration of object moving along a line is always 0, then its velocity is constant.
(C) It is impossible for the instantaneous velocity at all times a(D) A moving object can have negative acceleration and increasing speed.

Answers

Answer:

Explanation:(A)if a body is accelerating then it's velocity can't be constant since an object is said to be accelerating if it is changing velocity (B)if the acceleration of an object moving along a line is 0 then it's velocity will be constant since there is no change in direction or speed(C)No.it is not possible for a moving body to have an instantaneous velocity at all times since instantaneous velocity is the velocity of a body at a certain instant of time..(D)Yes a moving object can have a negative acceleration and increasing speed,it can also have a positive acceleration with decreasing speed.

Select True or False for the following statements about Heisenberg's Uncertainty Principle. True False It is not possible to measure simultaneously the x and y positions of a particle exactly.True False It is not possible to measure simultaneously the x and y momentum components of a particle exactly.
True False It is not possible to measure simultaneously the z position and the z momentum component of a particle exactly.

Answers

Answer:

Statement 1) False

Statement 2) False

Statement 3) True

Explanation:

The uncertainty principle states that " in a physical system certain quantities cannot be measured with random precision no matter whatever the least count of the instrument is" or we can say while measuring simultaneously the position and momentum of a particle the error involved is

$P\cdot\delta x\geq (h)/(4\pi )$

Thus if we measure x component of momentum of a particle with 100% precision we cannot measure it's position 100% accurately as the error will be always there.

Statement 1 is false since measurement of x and y positions has no relation to uncertainty.

Statement 2 is false as both the momentum components can be measured with 100% precision.

Statement 3 is true as as demanded by uncertainty principle since they are along same co-ordinates.

A traffic light weighing 200N hangs from a vertical cable tied to two other cables that are fastened to to a support ,as shown . The upper cables make angles 41° and 63° with the horizontal . Calculate the tension in of the three cables

Answers

Answer:

93.6 N in the 41° cable
155.6 N in the 63° cable
200 N in the vertical cable

Explanation:

Let T and U represent the tensions in the 41° and 63° cables, respectively. In order for the system to be stationary, the horizontal components of these tensions must balance, and the vertical components of these tensions must total 200 N.

Tcos(41°) =Ucos(63°) . . . . . balance of horizontal components

U = Tcos(41°)/cos(63°) . . . . write an expression for U

The vertical components must total 200 N, so we have ....

Tsin(41°) +Usin(63°) = 200

Tsin(41°) +Tcos(41°)sin(63°)/cos(63°) = 200

T(sin(41°)cos(63°) +cos(41°)sin(63°))/cos(63°) = 200

T = 200cos(63°)/sin(41° +63°) ≈ 93.6 . . . newtons

U = 200cos(41°)/sin(41° +63°) ≈ 155.6 . . . newtons

The vertical cable must have sufficient tension to balance the weight of the traffic light, so its tension is 200 N.

Then the tensions in the 3 cables are ...

41°: 93.6 N

63°: 155.6 N

90°: 200 N

The tension in each of the three cables are 94.29, 155.56 and 200 Newton respectively.

Given the following data:

Force = 200 Newton.

Angle 1 = 41°

Angle 2 = 63°

How to calculate the tension.

First of all, we would determine the third tension force based on the vertical component as follows:

$\sum F_y = 0\n\nT_3 - F_g =0\n\nT_3 - F_g=200\;N$

Next, we would apply Lami's theorem to resolve the forces acting on the traffic light at equilibrium:

For the horizontal component:

$\sum F_x = -T_1cos41+T_2cos 63=0\n\n0.7547T_1=0.4540T_2\n\nT_1=(0.4540T_2)/(0.7547)\n\nT_1 = 0.6016T_2$ ....equation 1.

For the vertical component:

$\sum F_y = T_1sin41+T_2sin 63-T_3=0\n\n\sum F_y = T_1sin41+T_2sin 63-200=0\n\n0.6561T_1+0.8910T_2 =200$ ...equation 2.

Substituting eqn. 1 into eqn. 2, we have:

$0.6561 * (0.6016T_2)+0.8910T_2 =200\n\n0.3947T_2+0.8910T_2 =200\n\n1.2857T_2 =200\n\nT_2 = (200)/(1.2857) \n\nT_2 = 155.56\;Newton$

For the first tension:

$T_1 = 0.6061T_2\n\nT_1 = 0.6061 * 155.56\n\nT_1 = 94.29\;Newton$

Read more on tension here: brainly.com/question/4080400

A popular physics lab involves a hand generator and an assortment of wires with different values of resistance. In the lab, the leads of the generator are connected across each wire in turn. For each wire, students attempt to turn the generator handle at the same constant rate. Students must push harder on the handle when the leads of the generator are connected__________. This is because turning the handle at a given constant rate produces__________ , regardless of what is connected to the leads. So, when turning the handle at a constant rate, lab students must push harder in cases where there is________

Answers

Answer:

Explanation:

Students must push harder on the handle when the leads of the generator are connected across the wire with the lowest resistance.

This is because turning the handle at a given constant rate produces a constant voltage across the leads, regardless of what is connected to the leads.

So, when turning the handle at a constant rate, lab students must push harder in case where there is a greater current through the connected wire.

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